Problem: Evaluate $~~\int^t_{0} x\sqrt{x-t}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{4}{15}(-t)^{5/2}$ (Choice B) B $(-t)^{5/2}$ (Choice C) C $\dfrac{2}{3}(-t)^{3/2}+\dfrac{4}{15}(-t)^{5/2}$ (Choice D) D $\dfrac{2}{3}(-t)^{5/2}$
Explanation: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = x~$ and $~dv=\sqrt{x-t}\ dx\,$. Then $~du = dx~$ and $~v = \dfrac23(x-t)^{3/2}$. Integration by parts gives $ \int^t_{0} x\sqrt{x-t}\,dx = x\cdot\dfrac23(x-t)^{3/2}\Bigg|_0^t-\int^t_{0}\dfrac23(x-t)^{3/2}dx$ $ ~=x\cdot\dfrac23(x-t)^{3/2}-\dfrac23\cdot\dfrac25(x-t)^{5/2}\Bigg]^t_{0}$ $ ~=x\cdot\dfrac23(x-t)^{3/2}-\dfrac4{15}(x-t)^{5/2}\Bigg]^t_{0}$ $ ~=0-\Big(0-\dfrac4{15}(-t)^{5/2}\Big)=\dfrac{4}{15}(-t)^{5/2}$.